NEET Chemistry Questions: Alcohol, phenol & ether

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Choose the proper option for given statement on the basis of physical properties Statement : (i) Alkyl isocyanides have bad odours while alkyl cyanides have pleasant odours. Statement : (ii) Alkyl cynanides are poisonous compounds. Statement : (iii) The boiling points of alkyl cyanides are lower than their isomeric alkyl-isocynides. Statement : (iv) Acetonitrile is soluble in water but methylcarbylamine is not.




No explanation available.
Inter molecular hydrogen bonding is strongest in




Intermolecular hydrogen bonding is strongest in methanol compared to methylamine, phenol, and methanal. This is because methanol (CH3OH) has a hydroxyl group (-OH) which can form strong hydrogen bonds with other methanol molecules. The oxygen atom in the hydroxyl group is highly electronegative and can attract hydrogen atoms from neighboring molecules, forming strong hydrogen bonds.
Among the following dissociation constant is highest for




$ CH_3 NH_3 ^+ Cl ^ -$ being asalt, undergoes almostcomplete dissociation, therefore, it has ahigh dissociation constant.
Numbers of isomeric alcohols of molecular formula $C_5H_{12}O $ are




For the molecular formula $C_5H_{12}O$, there are 8 possible isomeric alcohols. These include different structural isomers with variations in the position of the hydroxyl group ($-OH$) and branching of the carbon chain. The isomers can be primary, secondary, or tertiary alcohols, each with a unique structure. Therefore, the correct answer is 8.
Which of the following will produce only one product on reduction with $LiAlH_4$ ?




No explanation available.
$CH_3 CH_2 CH_2OH $ Can be converted to $CH_3 CH_2 CH_2COOH$ by the following sequence of steps:




The conversion of $CH_3CH_2CH_2OH$ (1-propanol) to $CH_3CH_2CH_2COOH$ (butanoic acid) involves the following steps: first, reaction with $PBr_3$ to replace the hydroxyl group with a bromine atom to form $CH_3CH_2CH_2Br$ (1-bromopropane). Next, this compound reacts with $KCN$ to form $CH_3CH_2CH_2CN$ (butyronitrile). Finally, hydrolysis of the nitrile with $H_3O^+$ yields the carboxylic acid $CH_3CH_2CH_2COOH$.
In the following Sequence of reactions,$ CH_3 CH_2 OH \xrightarrow [ ] { {P + I_2 }} A \xrightarrow [ { ether} ] { { Mg }} B \xrightarrow [ ] { {HCHO} } C \xrightarrow [ ] { {H_2 O } }D $ , the compound D is:




No explanation available.
Acid catalysed hydration of alkenes except ethene leads to the formation of




No explanation available.

During dehydration of alcohol to alkenes by heating with Conc $H_2SO_4$ , the initiation step is:





In the dehydration of alcohols to alkenes using concentrated $H_2SO_4$, the initiation step is the protonation of the alcohol molecule. This step involves the alcohol accepting a proton from the $H_2SO_4$, forming an oxonium ion, which makes the -OH group a better leaving group.
which of the following compounds will give positive iodoform test ?(I) 3- methyl propan-2-ol (II) I - methyl cyclopentanal (III ) I - phenyl propan-1-ol (Iv) 3- phenyl propan-2-ol.




The iodoform test is used to identify compounds with the structure $CH_3C(OH)R$ or $CH_3COR$. Compound (I) 3-methyl propan-2-ol and compound (IV) 3-phenyl propan-2-ol both have the $CH_3C(OH)R$ group, and will therefore give a positive iodoform test.
Propan -l-ol and propan-2-ol can be distinguished by




No explanation available.
$ CH_3CH_2OH $ cannot be prepare by which of the following reaction ?




Ethanol ($CH_3CH_2OH$) cannot be prepared by the reaction of ethyl chloride with alcoholic potassium hydroxide. This reaction typically results in the formation of ethene through an elimination reaction (dehydrohalogenation), not ethanol.
The most suitable reagent for the conversion of primary alcohol into aldehyde with the same number of carban is




Pyridinium chlorochromate (PCC) is a mild oxidizing agent that converts primary alcohols into aldehydes without further oxidation to carboxylic acids, which makes it the most suitable reagent for this conversion.
An organic compound "X" on treatment with PDC in $CH_2Cl_2 $ gives compound "Y". Compound "Y", reacts with $ I_2 $ and alkali to form yellow precipitate. The compound "X" is




The compound 'X' is ethanol. When ethanol is treated with Pyridinium Dichromate (PDC) in $CH_2Cl_2$, it gets oxidized to ethanal (compound 'Y'). Ethanal reacts with $I_2$ and alkali to form iodoform (yellow precipitate), confirming the presence of a methyl ketone or an aldehyde with a methyl group adjacent to the carbonyl group.
How many optically active stereoisomers are pos idle for butan -2,3 -diol?




Butane-2,3-diol has two chiral centers and can exist in three optically active stereoisomers: (2R,3R), (2S,3S), and one meso compound (2R,3S or 2S,3R) which is optically inactive. Therefore, only two optically active stereoisomers are possible.